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[(23 - 5) . (-3)+9] . (22012 . 2011 - 20122 . 2011+1)
= [ 3 . ( -3 ) + 9] . (22012 . 2011 - 20122 . 2011+1)
= [ (-9) + 9 ] . (22012 . 2011 - 20122 . 2011+1)
= 0 . (22012 . 2011 - 20122 . 2011+1)
= 0
\(A=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2011.2013}\right)\)
\(A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{4048144}{2011.2013}\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2012.2012}{2011.2013}\)
\(A=\frac{2.3.4...2012}{1.2.3...2011}.\frac{2.3.4...2012}{3.4.5...2013}\)
\(A=2012.\frac{2}{2013}=\frac{4024}{2013}\)
Ta có :
\(A=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right).................\left(1+\dfrac{1}{2011.2013}\right)\)
\(A=\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\left(\dfrac{8}{8}+\dfrac{1}{8}\right)................\left(\dfrac{9999}{9999}+\dfrac{1}{9999}\right)\)
\(A=\dfrac{4}{3}.\dfrac{9}{8}...............\dfrac{10000}{9999}\)
\(A=\dfrac{4.9.................10000}{3.8.............9999}\)
\(A=\dfrac{2.2.3.3................100.100}{1.3.2.4...............99.101}\)
\(A=\dfrac{2.100}{101}=\dfrac{200}{101}\)
~ Chúc bn học tốt ~
A= (1+1/1 x 3)x(1+1/2x4)x(1+1/3x5)x............x(1+1/2011x2013)
\(=\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{8}{8}+\frac{1}{8}\right)....\left(\frac{4048143}{4048143}+\frac{1}{4048143}\right)\)
\(=\frac{4}{3}\cdot\frac{9}{8}\cdot...\cdot\frac{4048144}{4048143}\)
\(=\frac{4\cdot9\cdot....\cdot4048144}{3\cdot8\cdot....\cdot4048143}\)
\(=\frac{2\cdot2\cdot3\cdot3\cdot....\cdot2012\cdot2012}{1\cdot3\cdot2\cdot4\cdot....\cdot2011\cdot2013}\)
\(=\frac{2\cdot2012}{2013}=\frac{4024}{2013}\)
ta có
(2011+1)+(2013-1)
=(2012-1)
vì (2012-1)<20122
=>2011x2013<20122
Ta có :
2011 . 2013
= 2011 . ( 2012 + 1 )
= 2011 . 2012 + 2011 . 1
20122
= 2012 . 2012
= ( 2011 + 1 ) . 2012
= 2011 . 2012 + 2012 . 1
Ta thấy : 2011 . 1 < 2012 . 1 nên 2011 . 2012 + 2011 . 1 < 2011 . 2012 + 2012 . 1
Vậy : 2011 . 2013 < 20122